from the definitions of Functional Derivatives used by mathematicians (I´ve seen many claims that it is, in effect, the Fréchet derivative, but no proofs). The Wikipedia article says it´s just a matter of using the delta function as “test function” but then goes on to say that it is nonsense.

Waiting for an actual answer, but note that Dirac delta comes from the fact that $F[x]$ is the functional applied to the function $x$ and the functional derivative is with respect to the function $y$
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JorgeJun 18 '12 at 23:07

4 Answers
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Whenever I have troubles with functional derivative things, I just do the replacement of a continuous variable $x$ into a discrete index $i$. If I'm not mistaken this is what they call a "DeWitt notation".

The hand waiving idea is that you can think of a functional $F[f(x)]$ as of a "ordinary function" of many variables $F(f_{-N},\cdots,f_0,f_1,f_2,\cdots,f_N) = F(\vec{f})$ with $N$ going to "continuous infinity".

In that language your functional derivative transforms into partial derivative over one of the variables:
$$\frac{\delta F}{\delta f(x)} \to \frac{\partial F}{\partial f_i}$$
And the delta-function is just an ordinary Kronecker delta:
$$\delta(x-y) \to \delta_{ij}$$

Where $\delta f$ is the infinitesimal change in f, and it is a smooth test function, and then on the right hand side, $A(x)$ is just a linear operator on the space of functions. The notation for the $A(x)$ is then

$$ A(x) = {\delta F\over \delta f(x)}$$

Because if you formally substitude $\delta f(x) = \delta(x-y)$, you find $A(y)$ as the value of the integral. This is just a notational trick--- $\delta f$ is an everywhere small variation, which is impossible if it is infinite at one point. Another way of saying this is that the point-delta-function limit has to be taken after the small epsilon limit in the definition you give, so that the variation becomes small before it becomes infinitely concentrated.

Obviously, the delta function isn't actually a function. But this use of it makes exactly as much sense as the position basis (and the latter can be made perfectly rigorous using rigged Hilbert spaces).

That´s exactly my problem with this definition (Dirac delta subtleties apart). It seems your are defining the partial derivative in the specific direction of $\delta_y$, but this very definition is used everywhere as the partial derivative in any direction (are they all the same?). Another way of expressing my problem: is the first equality below valid? Why? (I tried to follow your notation, $f_y$ means f calculated at point $y$): $$ \frac{\delta F}{\delta (f_y)}[f] = \frac{\delta F}{\delta (\delta_y)}[f] = \lim_{\epsilon \to 0} \frac{1}{\epsilon} ( F[f + \epsilon \delta_y] - F[f]). $$
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Forever_a_NewcomerJun 19 '12 at 18:10

The definition works for any function. You define the derivative of $F$ at $f$ in the direction of $f+g$ for any function $g$ to be $\frac{\delta F}{\delta g} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} (F[f+\epsilon g] - F[f])$. This is just like in multivariable calculus; it tells you how $F$ changes to first order if you move from $f$ towards $f+g$. The special degenerate case $g = \delta_y$ tells you how $F$ changes if you modify $f$ only by changing its value at $y$. This is the source of the crazy physics notation.
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user1504Jun 19 '12 at 22:49

Also, please don't use the notation $f_y$ for $f(y)$; this does not make the world a better place. $f(y)$ is the value of a function. $\delta_y$ (which seems to have inspired you) is a distribution which is non-zero only at $y$. If it helps, you can imagine that when physicists write $f(y)$ in the denominator of a functional derivative, they are actually taking the derivative along the "coordinate" $f \mapsto f(y)$.
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user1504Jun 19 '12 at 22:54